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Copy file name to clipboardExpand all lines: databases/catdat/data/categories/CompHaus.yaml
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reason: By the Tychonoff product theorem, a product in $\Top$ of compact Hausdorff spaces is compact; it is also clearly Hausdorff. Since the forgetful functor from $\CompHaus$ to $\Top$ is fully faithful, this limit is reflected in $\CompHaus$ as well.
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- property_id: equalizers
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reason: The equalizer in $\Top$ of two continuous functions $f, g : X \rightrightarrows Y$ between compact Hausdorff spaces is a closed subspace of $X$, and therefore it is also compact Hausdorff. Since the forgetful functor from $\CompHaus$ to $\Top$ is fully faithful, this limit is reflected in $\CompHaus$ as well.
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reason: 'The equalizer in $\Top$ of two continuous functions $f, g : X \rightrightarrows Y$ between compact Hausdorff spaces is a closed subspace of $X$, and therefore it is also compact Hausdorff. Since the forgetful functor from $\CompHaus$ to $\Top$ is fully faithful, this limit is reflected in $\CompHaus$ as well.'
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- property_id: cocomplete
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reason: $\CompHaus$ is a reflective subcategory of $\Top$, with the reflector being the Stone-Čech compactification functor. See <a href="https://ncatlab.org/nlab/show/compact+Hausdorff+space#StoneCechCompactification" target="_blank">nLab</a> for example. Therefore, as usual, we can form colimits in $\CompHaus$ by forming colimits in $\Top$ and then applying Stone-Čech compatification.
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reason: The forgetful functor from $\CompHaus$ to $\Set$ is monadic; see for example <a href="https://ncatlab.org/nlab/show/compact+Hausdorff+space#compact_hausdorff_spaces_are_monadic_over_sets">nLab</a>. Therefore, by <a href="https://ncatlab.org/nlab/show/colimits+in+categories+of+algebras#exact">this result</a>, $\CompHaus$ is Barr-exact, and in particular it has effective congruences.
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- property_id: cogenerator
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reason: The unit interval $[0, 1]$ is a cogenerator: Suppose we have $f, g : X \rightrightarrows Y$ with $f \ne g$. Choose $x\in X$ such that $f(x) \ne g(x)$. Then by Urysohn's lemma, there is a continuous function $h : Y \to [0, 1]$ such that $h(f(x)) = 0$ and $h(g(x)) = 1$. Therefore, $h\circ f \ne h\circ g$.
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reason: 'The unit interval $[0, 1]$ is a cogenerator: Suppose we have $f, g : X \rightrightarrows Y$ with $f \ne g$. Choose $x\in X$ such that $f(x) \ne g(x)$. Then by Urysohn''s lemma, there is a continuous function $h : Y \to [0, 1]$ such that $h(f(x)) = 0$ and $h(g(x)) = 1$. Therefore, $h\circ f \ne h\circ g$.'
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- property_id: extensive
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reason: This follows as for $\Top$ or $\Haus$ since finite coproducts in $\CompHaus$ are foemd as disjoint union spaces with the disjoint union topology.
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- property_id: epi-regular
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reason: |-
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First, any epimorphism $f : X\to Y$ is surjective: if not, its image would be a proper subset of $Y$, which is compact and hence closed. Then by Urysohn''s lemma, there would be a non-zero continuous function $g : Y \to [0, 1]$ which is $0$ on the image; but then $g \circ f = 0 \circ f$, giving a contradiction.
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First, any epimorphism $f : X\to Y$ is surjective: if not, its image would be a proper subset of $Y$, which is compact and hence closed. Then by Urysohn's lemma, there would be a non-zero continuous function $g : Y \to [0, 1]$ which is $0$ on the image; but then $g \circ f = 0 \circ f$, giving a contradiction.
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Now the identity morphism from $Y$, with the quotient topology of $f$, to $Y$ with its given topology is a bijective continuous function between compact Hausdorff spaces, so it is a homeomorphism. In other words, $f$ is a quotient map. Therefore, we see that if $g, h : E \rightrightarrows X$ is the kernel pair of $f$, and $U : \CompHaus \to \Top$ is the forgetful functor, then $U(f)$ is the coequalizer of $U(g)$ and $U(h)$. Since $U$ is fully faithful, that implies $f$ is the coequalizer of $g$ and $h$.
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- property_id: semi-strongly connected
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reason: Every non-empty compact Hausdorff space is weakly terminal (by using constant maps).
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reason: This is already true for <a href="/category/Top">$\mathbf{Top}$</a>.
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- property_id: coregular
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reason: It suffices to show that pushouts preserve (regular) monomorphisms in $\CompHaus$. Thus, suppose we have a pushout square
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reason: 'It suffices to show that pushouts preserve (regular) monomorphisms in $\CompHaus$. Thus, suppose we have a pushout square
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$$\begin{CD}
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A @> i >> B \\
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@V f VV @VV g V \\
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C @>> j > D,
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\end{CD}$$
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with $i : A \hookrightarrow B$ a monomorphism. Then for any pair of distinct elements $c, c' \in C$, by Urysohn's lemma there exists $\gamma : C \to [0, 1]$ with $\gamma(c) = 0$ and $\gamma(c') = 1$. Also, by Tietze's extension theorem, there exists $\beta : B \to [0, 1]$ such that $\beta \circ i = \gamma \circ f$. By the pushout property, there is a unique $\delta : D \to [0, 1]$ such that $\delta \circ g = \beta$ and $\delta \circ j = \gamma$. Since $\delta(j(c)) \ne \delta(j(c'))$, we conclude that $j(c) \ne j(c')$. This shows that $j$ is injective, so it is a regular monomorphism.
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with $i : A \hookrightarrow B$ a monomorphism. Then for any pair of distinct elements $c, c'' \in C$, by Urysohn''s lemma there exists $\gamma : C \to [0, 1]$ with $\gamma(c) = 0$ and $\gamma(c'') = 1$. Also, by Tietze''s extension theorem, there exists $\beta : B \to [0, 1]$ such that $\beta \circ i = \gamma \circ f$. By the pushout property, there is a unique $\delta : D \to [0, 1]$ such that $\delta \circ g = \beta$ and $\delta \circ j = \gamma$. Since $\delta(j(c)) \ne \delta(j(c''))$, we conclude that $j(c) \ne j(c'')$. This shows that $j$ is injective, so it is a regular monomorphism.'
reason: Suppose we have a cofiltered diagram of epimorphisms $(f_i : X_i \to Y_i)$, and $y = (y_i) \in \lim_i Y_i$. Then by <a href="/lemma/cofiltered-limit-of-non-empty-compact">this result</a>, the limit of $f_i^{-1}(\{ y_i \})$ is non-empty. If $x$ is in this limit, that implies that $(\lim_i f_i)(x) = y$.
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reason: 'Suppose we have a cofiltered diagram of epimorphisms $(f_i : X_i \to Y_i)$, and $y = (y_i) \in \lim_i Y_i$. Then by lemma 1 <a href="/pdf/comphaus_copresentable.pdf">here</a>, the limit of $f_i^{-1}(\{ y_i \})$ is non-empty. If $x$ is in this limit, that implies that $(\lim_i f_i)(x) = y$.'
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- property_id: locally copresentable
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reason: A proof can be found <a href="/pdf/comphaus_copresentable.pdf">here</a>.
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unsatisfied_properties:
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- property_id: Malcev
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reason: This is clear since $\FinSet$ is not Malcev and can be interpreted as the subcategory of finite discrete spaces
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reason: This is clear since $\FinSet$ is not Malcev and can be interpreted as the subcategory of finite discrete spaces.
reason: The proof is similar to <a href="/category/Haus">$\Haus$</a>. For $n \geq 1$ let $X_n$ be the pushout of $[1/n, 1] \hookrightarrow [0, 1]$ with itself. That is, $X_n$ is the union of two unit intervals $[0, 1] \times \{ 1 \}$ and $[0, 1] \times \{ 2 \}$ where we identify $(x,1) \equiv (x,2)$ when $x \geq 1/n$. As in the construction for $\Haus$, we see that the colimit in $\Haus$ is $[0, 1]$ where all corresponding points of both unit intervals are identified. Since this is compact Hausdorff, it also provides the colimit in $\CompHaus$. Again, the injective continuous maps $\{1,2\} \to X_n$, $i \mapsto (0,i)$ (where $\{1,2\}$ is discrete) become the constant map $0 : \{1,2\} \to [0,1]$ in the colimit, which is not a monomorphism.
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reason: 'The proof is similar to <a href="/category/Haus">$\Haus$</a>. For $n \geq 1$ let $X_n$ be the pushout of $[1/n, 1] \hookrightarrow [0, 1]$ with itself. That is, $X_n$ is the union of two unit intervals $[0, 1] \times \{ 1 \}$ and $[0, 1] \times \{ 2 \}$ where we identify $(x,1) \equiv (x,2)$ when $x \geq 1/n$. As in the construction for $\Haus$, we see that the colimit in $\Haus$ is $[0, 1]$ where all corresponding points of both unit intervals are identified. Since this is compact Hausdorff, it also provides the colimit in $\CompHaus$. Again, the injective continuous maps $\{1,2\} \to X_n$, $i \mapsto (0,i)$ (where $\{1,2\}$ is discrete) become the constant map $0 : \{1,2\} \to [0,1]$ in the colimit, which is not a monomorphism.'
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- property_id: exact cofiltered limits
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Consider the $\IN$-codirected systems $X_n := [0, 1] \times [0, 1/n]$ with the maps $X_{n+1} \to X_n$ being inclusion maps, and $Y_n := [0, 1+1/n]$ with the maps $Y_{n+1} \to Y_n$ also being inclusion maps. We define $f_n : X_n \to Y_n$, $(x, y) \mapsto x$ and $g_n : X_n \to Y_n$, $(x, y) \mapsto x+y$. It is straightforward to check these give morphisms of $\IN$-codirected systems in $\CompHaus$.
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Now for each $n$, the coequalizer of $f_n$ and $g_n$ is a single-point space. On the other hand, $\lim X_n \simeq [0, 1] \times \{ 0 \}$; $\lim Y_n \simeq [0, 1]$; and $\lim f_n = \lim g_n$, $(x, 0) \mapsto x$. Thus, the coequalizer of $\lim f_n$ and $\lim g_n$ is $[0, 1]$, showing that this coequalizer is not preserved under limits.'
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Now for each $n$, we claim the coequalizer of $f_n$ and $g_n$ is a singleton space. To see this, we prove the more general result that for $r, s > 0$ the coequalizer of $f, g : [0, r] \times [0, s] \rightrightarrows [0, r+s]$, $f(x,y) = x$, $g(x,y) = x+y$ is a singleton. We must show that for any $h : [0, r+s] \to T$ with $h\circ f = h\circ g$, then $h$ is constant. To this end, we show by induction on $n$ that whenever $x \in [0, r+s]$ and $x \le ns$, we have $h(x) = h(0)$. The base case $n=0$ is trivial. For the inductive step, if $x \le s$, then $f(0,x) = 0$ and $g(0,x) = x$, so $h(0) = h(x)$. Otherwise, we have $x-s \in [0,r]$ and $x-s \le (n-1)s$, so by inductive hypothesis $h(x-s) = h(0)$. Also, $f(x-s, s) = x-s$ and $g(x-s, s) = x$, so $h(x-s) = h(x)$, completing the induction. With this established, the desired result follows from the case $n := \lceil r/s \rceil + 1$.
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On the other hand, $\lim X_n \simeq [0, 1] \times \{ 0 \}$; $\lim Y_n \simeq [0, 1]$; and $\lim f_n = \lim g_n$, $(x, 0) \mapsto x$. Thus, the coequalizer of $\lim f_n$ and $\lim g_n$ is $[0, 1]$, showing that the limit does not preserve this coequalizer.
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special_objects:
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initial object:
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reason: This is easy.
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monomorphisms:
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description: injective continuous maps (which are automatically closed embeddings)
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reason: For the non-trivial direction, the forgetful functor to $\Set$ is representable (by the terminal object), hence preserves monomorphisms. To prove the parenthetical remark, given an injective continuous function $f : X \to Y$ between compact Hausdorff spaces, the image of $f$ is a closed subset. Also, the induced map from $X$ to $\im(f)$ with the subspace topology is a bijective continuous map between compact Hausdorff spaces, so it is a homeomorphism.
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reason: 'For the non-trivial direction, the forgetful functor to $\Set$ is representable (by the terminal object), hence preserves monomorphisms. To prove the parenthetical remark, given an injective continuous function $f : X \to Y$ between compact Hausdorff spaces, the image of $f$ is a closed subset. Also, the induced map from $X$ to $\im(f)$ with the subspace topology is a bijective continuous map between compact Hausdorff spaces, so it is a homeomorphism.'
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epimorphisms:
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description: surjective continuous maps (which are automatically quotient maps)
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reason: For the non-trivial direction, and for a proof of the parenthetical remark, see the proof above that $\CompHaus$ is epi-regular.
abstract = {Investigating dual local presentability of some topological and uniform classes, a new procedure is developed for factorization of maps defined on subspaces of products and a new characterization of local presentability is produced. The factorization is related to large cardinals and deals, mainly, with realcompact spaces. Instead of factorization of maps on colimits, local presentability is characterized by means of factorization on products.}
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}
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@article{Marra_2017,
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title={Stone duality above dimension zero: Axiomatising the algebraic theory of {C(X)}},
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